Solve a^2+4a^2+2a+1=0

$$a ^ { 2 } + 4 a ^ { 2 } + 2 a + 1 = 0$$

Solve for a

$$a=-\frac{1}{5}+\frac{2}{5}i=-0.2+0.4i$$

$$a=-\frac{1}{5}-\frac{2}{5}i=-0.2-0.4i$$

Steps Using the Quadratic Formula

Combine a^{2} and 4a^{2} to get 5a^{2}.

$$5a^{2}+2a+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-2±\sqrt{2^{2}-4\times 5}}{2\times 5}$$

Square 2.

$$a=\frac{-2±\sqrt{4-4\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$a=\frac{-2±\sqrt{4-20}}{2\times 5}$$

Add 4 to -20.

$$a=\frac{-2±\sqrt{-16}}{2\times 5}$$

Take the square root of -16.

$$a=\frac{-2±4i}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{-2±4i}{10}$$

Now solve the equation a=\frac{-2±4i}{10} when ± is plus. Add -2 to 4i.

$$a=\frac{-2+4i}{10}$$

Divide -2+4i by 10.

$$a=-\frac{1}{5}+\frac{2}{5}i$$

Now solve the equation a=\frac{-2±4i}{10} when ± is minus. Subtract 4i from -2.

$$a=\frac{-2-4i}{10}$$

Divide -2-4i by 10.

$$a=-\frac{1}{5}-\frac{2}{5}i$$

The equation is now solved.

$$a=-\frac{1}{5}+\frac{2}{5}i a=-\frac{1}{5}-\frac{2}{5}i$$

Steps for Completing the Square

Combine a^{2} and 4a^{2} to get 5a^{2}.

$$5a^{2}+2a+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$5a^{2}+2a=-1$$

Divide both sides by 5.

$$\frac{5a^{2}+2a}{5}=-\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$a^{2}+\frac{2}{5}a=-\frac{1}{5}$$

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{2}{5}a+\left(\frac{1}{5}\right)^{2}=-\frac{1}{5}+\left(\frac{1}{5}\right)^{2}$$

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{2}{5}a+\frac{1}{25}=-\frac{1}{5}+\frac{1}{25}$$

Add -\frac{1}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{2}{5}a+\frac{1}{25}=-\frac{4}{25}$$

Factor a^{2}+\frac{2}{5}a+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{1}{5}\right)^{2}=-\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{1}{5}\right)^{2}}=\sqrt{-\frac{4}{25}}$$

Simplify.

$$a+\frac{1}{5}=\frac{2}{5}i a+\frac{1}{5}=-\frac{2}{5}i$$

Subtract \frac{1}{5} from both sides of the equation.

$$a=-\frac{1}{5}+\frac{2}{5}i a=-\frac{1}{5}-\frac{2}{5}i$$